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[parent] examples of Einstein summation notation (Example)

Some examples of applying the Einstein summation notation.




Example 1. Let us consider the quantity

$\displaystyle S = a_{\alpha \beta} x^{\alpha}x^{\beta}$

for a three dimensional space. Since the index $\alpha$ occurs as both a subscript and a superscript, we sum on $\alpha$ from 1 to 3. This yields

$\displaystyle S = a_{1 \beta}x^1x^{\beta} + a_{2 \beta}x^2x^{\beta} + a_{3 \beta}x^3 x^{\beta}$

Now each term of $S$ is such that $\beta$ is both a subscript and superscript. Summing on $\beta$ from 1 to 3 as prescribed by our summation convention yields the quadratic form


$\displaystyle S = a_{11}x^1x^{1} + a_{12}x^1x^2 + a_{13}x^1x^3$      
$\displaystyle + a_{21}x^2x^1 + a_{22}x^2x^2 + a_{23}x^2x^3$      
$\displaystyle + a_{31}x^3 x^1+ a_{32}x^3x^2 + a_{33}x^3x^3$      




Example 2. If $x^1, x^2, x^3, \dots, x^n$ is a set of independent variables, then

$\displaystyle \frac{\partial x^1}{\partial x^1} = \frac{\partial x^2}{\partial ... ...rac{\partial x^3}{\partial x^3} = \dots = \frac{\partial x^n}{\partial x^n} = 1$

and if $i \ne j$

$\displaystyle \frac{\partial x^1}{\partial x^2} = 0, \qquad \frac{\partial x^i}{\partial x^j} = 0$

We may write

\begin{displaymath}\frac{\partial x^i}{\partial x^j} \equiv \delta_j^i \begin{ca... ...quad i = j \ = & 0 \qquad \text{if} \quad i \ne j \end{cases}\end{displaymath} (1)

The symbol $\delta_j^i$ is called the Kronecker delta. We have

$\displaystyle \delta_\alpha^\alpha = \delta_1^1+\delta_2^2 + \cdots + \delta_n^n = n$

Let us now assume that the quadratic form at the end of example 1 vanishes identically for all values of the independent variables $x^1$,$x^2$, $x^3$, and $a_{ij}$ to be constant. Differentiating $S=a_{\alpha \beta}x^\alpha x^\beta = 0$ with respect to a given variable, say $x^i$, yields


$\displaystyle \frac{\partial S}{\partial x^i} = a_{\alpha \beta}x^\alpha \frac{... ...rtial x^i} + a_{\alpha \beta}x^\beta \frac{\partial x^\alpha}{\partial x^i} = 0$      
$\displaystyle \frac{\partial S}{\partial x^i} = a_{\alpha \beta}x^\alpha \delta_i^\beta + a_{\alpha \beta}x^\beta \delta_i^\alpha = 0$      
$\displaystyle \frac{\partial S}{\partial x^i} = a_{\alpha i}x^\alpha + a_{i\beta}x^\beta = 0$      

Now differentiating with respect to $x^i$ yields

$\displaystyle \frac{\partial^2 S}{\partial x^j \partial x^i} = a_{\alpha i} \delta_j^\alpha + a_{i \beta}\delta_i^\beta = 0$

so that $a_{ji} + a_{ij} = 0$ or $a_{ij} = -a_{ji}$ for $i,j = 1,2,3$.




Example 3. We define $\epsilon^{ij}, i, j = 1,2$, to have the following numerical values: Let $\epsilon^{11} = \epsilon^{22} = 0, \epsilon^{12} = 1, \epsilon^{21} = -1$. We now consider the expression

$\displaystyle D = \epsilon^{ij} a_i^1 a_j^2$ (2)

Expanding (2) by use of our summation convention yields

$\displaystyle D = \epsilon^{11}a_1^1a_1^2 + \epsilon^{12}a_1^1 a_2^2 + \epsilon^{21}a_2^1 a_1^2 + \epsilon^{22}a_2^1a_2^2 = a_1^1a_2^2 - a_2^1a_1^2 $

The reader who is familiar with second-order determinants quickly recognizes that

$\displaystyle \epsilon^{ij} a_i^1 a_j^2 = \left \vert \begin{array}{cc} a_1^1 & a_2^1 \ a_1^2 & a_2^2 \end{array} \right \vert$ (3)




Example 4. The system of equations

$\displaystyle \left ( \begin{array}{ccc} y^1 & = & y^1(x^1, x^2, \cdots, x^n) \... ...& \vdots & \vdots \ y^n & = & y^n(x^1, x^2, \cdots, x^n) \end{array} \right )$ (4)

represents a coordinate transformation from an $(x^1, x^2, \cdots, x^n)$ coordinate system to a $(y^1, y^2, \cdots, y^n)$ coordinate system. From the calculus we have

$\displaystyle dy^i = \frac{\partial y^i}{\partial x^1} dx^1 + \frac{\partial y^... ...}dx^2 + \cdots + \frac{\partial y^i}{\partial x^n} dx^n \qquad i = 1,2,\cdots,n$

$\displaystyle dy^i = \frac{\partial y^i}{\partial y^\alpha} d x^\alpha$

The $\alpha$ in the term $\frac{\partial y^i}{\partial x^\alpha} $ is to be considered as a subscript. If, furthermore, the $x^i$, $i = 1, 2, \cdots, n$, can be solved for the $y^1, y^2, \cdots, y^n$, and assuming differentiability of the $x^i$ with respect to each $y^i$, one obtains

$\displaystyle \frac{\partial y^i}{\partial y^j} \equiv \delta_j^i = \frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}$

Differentiating this expression with respect to $y^k$ yields

$\displaystyle 0 = \frac{\partial y^i}{\partial x^\alpha} \frac{\partial^2 x^\al... ...a} \frac{\partial x^\beta}{\partial y^k} \frac{\partial x^\alpha}{\partial y^j}$

Multiplying both sides of this equation by $\frac{\partial x^\sigma}{\partial y^i} $ amd summing on the inex $i$ yields

$\displaystyle 0 = \frac{\partial x^\sigma}{\partial y^i} \frac{\partial y^i}{\p... ...} \frac{\partial x^\alpha}{\partial y^j} \frac{\partial x^\sigma}{\partial y^i}$

or

$\displaystyle 0 = \delta_\alpha^\sigma \frac{\partial^2 x^\alpha}{\partial y^k ... ...} \frac{\partial x^\alpha}{\partial y^j} \frac{\partial x^\sigma}{\partial y^i}$

which yields

$\displaystyle \frac{\partial^2 x^\sigma}{\partial y^k \partial y^j} = - \frac{\... ...} \frac{\partial x^\alpha}{\partial y^j} \frac{\partial x^\sigma}{\partial y^i}$

In particular, if $y = f(x)$, then

$\displaystyle \frac{d^2x}{dy^2} = - \frac{d^2y}{dx^2}\left(\frac{dx}{dy} \right)^3$

References

[1] Lass, Harry. "Elements of pure and applied mathematics" New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public domain work [1].



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